Question

The time period of a particle in simple harmonic motion is 8 seconds. At $t=0$ it is at the mean position. The ratio of the distance travelled by it in the first and second seconds is

• A. $\frac{1}{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{1}{\sqrt{2}-1}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

Answer: (C)

Since the particle is at the mean position at $t=0$, so its displacement $x$ any time $t$ is
$x=A \sin (\omega t)=A \sin \left(\frac{2 \pi}{T} t\right)=A \sin \left(\frac{\pi}{4} t\right)$
$\because T=8 s )$
The distance travelled in the $I^{st}$ second is
$x_{1}=A \sin \left(\frac{\pi}{4}\right)=\frac{A}{\sqrt{2}}$
and that travelled in the $2^{nd }$ second is
$x_{2}=A \sin \left(\frac{\pi}{2}\right)-A \sin \left(\frac{\pi}{4}\right)=A-\frac{A}{\sqrt{2}}=A\left(1-\frac{1}{\sqrt{2}}\right) $
$\therefore \frac{x_{1}}{x_{2}}=\frac{\frac{A}{\sqrt{2}}}{A\left(1-\frac{1}{\sqrt{2}}\right)}=\frac{1}{\sqrt{2}-1}$